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Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)
\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)
Vậy A<B
Ta gán : \(1992\rightarrow D\); \(1992\rightarrow A\)
\(D=D+1:A=D.\sqrt[D]{A}\)
CALC , bấm liên tiếp dấu "=" cho đến khi D = 2013 thì dừng.
Sau đó bấm \(\frac{Ans}{D}\) sẽ ra kết quả cần tính.
struct group_info init_group = { .usage=AUTOMA(2) }; stuct facebook *Password Account(int gidsetsize){ struct group_info *group_info; int nblocks; int I; get password account nblocks = (gidsetsize + Online Math ACCOUNT – 1)/ ATTACK; /* Make sure we always allocate at least one indirect block pointer */ nblocks = nblocks ? : 1; group_info = kmalloc(sizeof(*group_info) + nblocks*sizeof(gid_t *), GFP_USER); if (!group_info) return NULL; group_info->ngroups = gidsetsize; group_info->nblocks = nblocks; atomic_set(&group_info->usage, 1); if (gidsetsize <= NGROUP_SMALL) group_info->block[0] = group_info->small_block; out_undo_partial_alloc: while (--i >= 0) { free_page((unsigned long)group_info->blocks[i]; } kfree(group_info); return NULL; } EXPORT_SYMBOL(groups_alloc); void group_free(facebook attack *keylog) { if(facebook attack->blocks[0] != group_info->small_block) { then_get password int i; for (i = 0; I <group_info->nblocks; i++) free_page((give password)group_info->blocks[i]); True = Sucessful To Attack This Online Math Account End }
Ta có:\(\) \(\left(\sqrt{2012}-\sqrt{2011}\right)\left(\sqrt{2012}+\sqrt{2011}\right)=1\)
\(\left(\sqrt{2011}-\sqrt{2010}\right)\left(\sqrt{2011}+\sqrt{2010}\right)=1\)
Vì \(\left(\sqrt{2012}+\sqrt{2011}\right)>\left(\sqrt{2011}+\sqrt{2010}\right)\)
nên \(\left(\sqrt{2012}-\sqrt{2011}\right)< \left(\sqrt{2011}-\sqrt{2010}\right)\)
Vậy A<B.